Problem: You have found the following ages (in years) of 4 zebras. Those zebras were randomly selected from the 48 zebras at your local zoo: $ 8,\enspace 13,\enspace 5,\enspace 10$ Based on your sample, what is the average age of the zebras? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we only have data for a small sample of the 48 zebras, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $4$ samples and divide by $4$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\overline{x}} = \dfrac{8 + 13 + 5 + 10}{{4}} = {9\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {1} + {16} + {16} + {1}} {{4 - 1}} $ {s^2} = \dfrac{{34}}{{3}} = {11.33\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{11.33\text{ years}^2}} = {3.4\text{ years}} $ We can estimate that the average zebra at the zoo is 9 years old. There is also a standard deviation of 3.4 years.